Problem: The four zeros of the polynomial $x^4 + jx^2 + kx + 225$ are distinct real numbers in arithmetic progression.  Compute the value of $j.$
Let the four roots be $a,$ $a + d,$ $a  + 2d,$ and $a + 3d.$  Then by Vieta's formulas, their sum is 0:
\[4a + 6d = 0.\]Then $d = -\frac{2}{3} a,$ so the four roots are $a,$ $\frac{a}{3},$ $-\frac{a}{3},$ and $-a.$  Their product is
\[a \cdot \frac{a}{3} \cdot \left( -\frac{a}{3} \right) (-a) = \frac{a^4}{9} = 225,\]so $a = \pm 3 \sqrt{5}.$  Hence, the four roots are $3 \sqrt{5},$ $\sqrt{5},$ $-\sqrt{5},$ $-3 \sqrt{5},$ and the polynomial is
\[(x - 3 \sqrt{5})(x - \sqrt{5})(x + \sqrt{5})(x + 3 \sqrt{5}) = (x^2 - 5)(x^2 - 45) = x^4 - 50x^2 + 225.\]Thus, $j = \boxed{-50}.$